{"id":28346,"date":"2025-06-20T11:54:04","date_gmt":"2025-06-20T11:54:04","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=28346"},"modified":"2025-06-20T11:54:05","modified_gmt":"2025-06-20T11:54:05","slug":"standard-form-of-a-quadratic-find-the-vertex-of-a-quadratic-function-fx-3x2-18x-32","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/standard-form-of-a-quadratic-find-the-vertex-of-a-quadratic-function-fx-3x2-18x-32\/","title":{"rendered":"Standard Form of a Quadratic Find the vertex of a quadratic function f(x) = 3x^2 + 18x + 32."},"content":{"rendered":"\n

Standard Form of a Quadratic Find the vertex of a quadratic function f(x) = 3x^2 + 18x + 32. Find the vertex, axis of symmetry, and the graph of f(x) = 4x^2 – 8x + 19. Vertex, Axis of Symmetry, and Graph.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Problem 1:<\/h3>\n\n\n\n

Find the vertex of the quadratic function<\/strong>
f(x)=3×2+18x+32f(x) = 3x^2 + 18x + 32f(x)=3×2+18x+32<\/p>\n\n\n\n

To find the vertex<\/strong> of a quadratic function in standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c, we use the formula for the x-value of the vertex:x=\u2212b2ax = \\frac{-b}{2a}x=2a\u2212b\u200b<\/p>\n\n\n\n

Here,
a=3a = 3a=3,
b=18b = 18b=18x=\u2212182\u22c53=\u2212186=\u22123x = \\frac{-18}{2 \\cdot 3} = \\frac{-18}{6} = -3x=2\u22c53\u221218\u200b=6\u221218\u200b=\u22123<\/p>\n\n\n\n

Now substitute x=\u22123x = -3x=\u22123 into the original function to find the y-value:f(\u22123)=3(\u22123)2+18(\u22123)+32=3(9)\u221254+32=27\u221254+32=5f(-3) = 3(-3)^2 + 18(-3) + 32 = 3(9) – 54 + 32 = 27 – 54 + 32 = 5f(\u22123)=3(\u22123)2+18(\u22123)+32=3(9)\u221254+32=27\u221254+32=5<\/p>\n\n\n\n

So the vertex<\/strong> is:(\u22123,5)\\boxed{(-3, 5)}(\u22123,5)\u200b<\/p>\n\n\n\n

\n\n\n\n

Problem 2:<\/h3>\n\n\n\n

Find the vertex, axis of symmetry, and graph the function<\/strong>
f(x)=4×2\u22128x+19f(x) = 4x^2 – 8x + 19f(x)=4×2\u22128x+19<\/p>\n\n\n\n

Again, use the vertex formula:x=\u2212b2a=\u2212(\u22128)2\u22c54=88=1x = \\frac{-b}{2a} = \\frac{-(-8)}{2 \\cdot 4} = \\frac{8}{8} = 1x=2a\u2212b\u200b=2\u22c54\u2212(\u22128)\u200b=88\u200b=1<\/p>\n\n\n\n

Substitute x=1x = 1x=1 into the function to find the y-value:f(1)=4(1)2\u22128(1)+19=4\u22128+19=15f(1) = 4(1)^2 – 8(1) + 19 = 4 – 8 + 19 = 15f(1)=4(1)2\u22128(1)+19=4\u22128+19=15<\/p>\n\n\n\n

So the vertex<\/strong> is:(1,15)\\boxed{(1, 15)}(1,15)\u200b<\/p>\n\n\n\n

The axis of symmetry<\/strong> is the vertical line that passes through the x-value of the vertex:x=1\\boxed{x = 1}x=1\u200b<\/p>\n\n\n\n

Summary and Explanation (300 words)<\/h3>\n\n\n\n

Quadratic functions in standard form are written as f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c. The shape of the graph is a parabola. If a>0a > 0a>0, the parabola opens upward. If a<0a < 0a<0, it opens downward. The vertex is the turning point of the parabola and is found using the formula x=\u2212b2ax = \\frac{-b}{2a}x=2a\u2212b\u200b. Once the x-value is found, plug it back into the function to find the corresponding y-value. This gives the coordinates of the vertex.<\/p>\n\n\n\n

In the first example, the function f(x)=3×2+18x+32f(x) = 3x^2 + 18x + 32f(x)=3×2+18x+32 has a vertex at (\u22123,5)(-3, 5)(\u22123,5). This point represents the minimum value of the function since the parabola opens upward. The second example, f(x)=4×2\u22128x+19f(x) = 4x^2 – 8x + 19f(x)=4×2\u22128x+19, has a vertex at (1,15)(1, 15)(1,15) and also opens upward because the coefficient of x2x^2×2 is positive. The axis of symmetry is simply a vertical line that passes through the x-value of the vertex. It divides the graph into two symmetrical halves.<\/p>\n\n\n\n

To graph either function, plot the vertex, draw the axis of symmetry, and then choose x-values on either side of the vertex to plot additional points. The symmetry of the graph helps to complete the shape of the parabola.<\/p>\n\n\n\n

These methods are useful in algebra and real-world applications where finding maximum or minimum values is important, such as in business or physics problems.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Standard Form of a Quadratic Find the vertex of a quadratic function f(x) = 3x^2 + 18x + 32. Find the vertex, axis of symmetry, and the graph of f(x) = 4x^2 – 8x + 19. Vertex, Axis of Symmetry, and Graph. The Correct Answer and Explanation is: Problem 1: Find the vertex of the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-28346","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28346","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=28346"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28346\/revisions"}],"predecessor-version":[{"id":28348,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28346\/revisions\/28348"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28346"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28346"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28346"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}