Graph the following quadratics and state its vertex, intercepts and domain & range.
$h (x) = \frac{1}{2} x^{2} + 10 x - 5$

Graph of a downward-opening parabola with vertex near (−10, −55) on an x-y coordinate plane.

Vertex: $$\left$(-10,-55$\right$)$ ; $x$ -intercepts: $$\left$(0.5,0$\right$),$\left$(-20.5,0$\right$)$ ; $y$ -intercept: $$\left$(0,-5$\right$)$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $$\left\[\lbrack$-55,$\infty\]\right$)$

Graph of a downward-opening parabola with vertex near (-10, -50) and x-intercepts around -20 and 0.

Vertex: $$\left$(10,55$\right$)$ ; $x$ -intercepts: $open paren negative 0.5 comma 0 close paren comma open paren 20.5 comma 0 close paren$ ; $y$ -intercept: $$\left$(0,-5$\right$)$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $$\left\[\lbrack$55,$\infty\]\right$)$

Graph of a downward-opening parabola with vertex near (−10, −55) and x-intercepts near (−20, 0) and (0, 0).

Graph of a downward-opening parabola with vertex near (−10, −55) on a coordinate grid.

Vertex: $open paren 10 comma 55 close paren$ ; $x$ -intercepts: $$\left$(-0.5,0$\right$),$\left$(20.5,0$\right$)$ ; $y$ -intercept: $open paren 0 comma negative 5 close paren$ ; Domain: $(- \infty, \infty)$ ; Range: $open bracket 55 comma infinity close paren$

Verified step by step guidance

Identify the quadratic function given: $h(x) = \frac{1}{2}x^2 + 10x - 5$.

Find the vertex using the vertex formula $x = -\frac{b}{2a}$, where $a = \frac{1}{2}$ and $b = 10$. Substitute this $x$ value back into $h(x)$ to find the $y$-coordinate of the vertex.

Calculate the $x$-intercepts by setting $h(x) = 0$ and solving the quadratic equation $\frac{1}{2}x^2 + 10x - 5 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Find the $y$-intercept by evaluating $h(0)$, which is simply the constant term in the quadratic expression.

State the domain and range: The domain of any quadratic function is all real numbers $\left(-\infty, \infty\right)$. The range depends on the vertex; since $a > 0$, the parabola opens upward, so the range is $\left[y_{vertex}, \infty\right)$.

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Graph the following quadratics and state its vertex, intercepts and domain & range.h(x)=12x2+10x−5h(x)=\(\frac\)12x^2+10x-5

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Graph the following quadratics and state its vertex, intercepts and domain & range.
$h (x) = \frac{1}{2} x^{2} + 10 x - 5$